I have to find the size and position of a window given four light points. The light points are i.i.d. from a uniform distribution with the parameter $\theta := (x_{min},x_{max},y_{min},y_{max})$. Each point is $(0,0),(0,1),(1,1),(2,2)$.
First I need to complete the pdf, $f_\theta (x,y)$, by finding the value c, through the parameters given.
I though this was quite easy, as I just used $\frac{1}{\Delta x \cdot \Delta y}=\frac{1}{(x_{max}-x_{min})\cdot (y_{max}-y_{min})}$
Next I have two values for $\theta$ that I need to calculate the likelihood for. $\theta_1 = (-1,4,-1,3)$ and $\theta_2 = (-2,5,-3,6)$. So we can calculate the value c for both of them: $c_1=\frac{1}{20}$, $c_2=\frac{1}{63}$. Here comes my first real question. Since it is a uniform distribution is it okay to just say:
$$\mathcal{L}_{\theta_1} = \prod_{i=1}^n \frac{1}{(4-(-1))\cdot(3-(-1))}=\prod_{i=1}^n \frac{1}{20}= \frac{1}{160,000}, \quad n=4$$
$$\mathcal{L}_{\theta_2} = \prod_{i=1}^n \frac{1}{(5-(-2))\cdot(6-(-3))}= \prod_{i=1}^n \frac{1}{63}= \frac{1}{15,752,961}, \quad n=4$$
Is this acceptable?
After that I need to calculate the $MLE\; \hat{\theta}^{ML}=(\hat{x}_{min},\hat{x}_{max},\hat{y}_{min},\hat{y}_{max})$. I am guessing the result would be a vector of the differentiated function for $x_{min}$ for $x_{max}$ for $y_{min}$ for $y_{max}$ equal to $0$? I'm guessing I should use the natural logaritmetic of the function as it is easier, but I'm stuck at what they expect me to do. If I differentiate $\prod_{i=1}^n ln(\frac{1}{(x_{max}-x_{min})\cdot (y_{max}-y_{min})})$ with respect to $x_{max}$ I get $\sum_{i=1}^n \frac{1}{x_{min}-x_{max}}$. I'll still have two variables. What am I missing? I think, I might have to work with
Best Answer
You can’t get the maximum likelihood estimate through differentiation in this case because it’s on the boundary. The density is higher the smaller the window is, so the window with maximal density (and hence likelihood) is the smallest one compatible with the observations. This is $\theta=(0,2,0,2)$.
Your likelihood calculation is correct, but you don’t mention the essential point that the four observations lie in both of these windows and that otherwise their likelihood would be $0$.
It’s not entirely clear what you mean by “find the pdf, $c$”. What you later calculate as $c$ is just the constant normalization factor in the density; the probability density function is the product of this constant factor with the indicator function of the window.